What is the general solution for 2cos^2(θ)-3cos(θ)+1=0,0<= θ<2pi ?

The general solution for 2cos^2(θ)-3cos(θ)+1=0,0<= θ<2pi is θ=0,θ= pi/3 ,θ=(5pi)/3

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2cos^2(θ)-3cos(θ)+1=0,0<= θ<2pi

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, 0 \leq θ < 2 π

θ = 0, θ = \frac{π}{3}, θ = \frac{5 π}{3}

Solution steps

2 cos^{2} (θ) - 3 cos (θ) + 1 = 0, 0 \leq θ < 2 π

Solve by substitution

cos (θ) = 1, cos (θ) = \frac{1}{2}

cos (θ) = 1, 0 \leq θ < 2 π : θ = 0

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π : θ = \frac{π}{3}, θ = \frac{5 π}{3}

Combine all the solutions

θ = 0, θ = \frac{π}{3}, θ = \frac{5 π}{3}

cos (x) sin (x) = 1

24 arctan (x) = 4 π

cos^{2} (x) - sin (x) - 1 = 0

5 cos (x) + 12 sin (x) = 6

- 73 tan (θ) + 5 = 26

What is the general solution for 2cos^2(θ)-3cos(θ)+1=0,0<= θ<2pi ?
The general solution for 2cos^2(θ)-3cos(θ)+1=0,0<= θ<2pi is θ=0,θ= pi/3 ,θ=(5pi)/3